If x^2y^3 = 200, what is the value of xy ?

Even though, the solution has already been provided above, I want to point out something. The silver lining of DS questions is that you don't need to solve. If you know you will get a unique value, you are done.

Question stem: If \(x^2y^3=200\), what is xy?

Statement 1: y is an integer but x needn't be.
If y = 1, \(x = +-\sqrt{200}\)
If y = 2, x = +-5 etc
We do not know what xy is.

Statement 2:\(\frac{x}{y} = \frac{5}{2}\)
This means \(x = (\frac{5}{2})y\)
This value of x you can put above to get a unique value of y. (Since you will get y^5 = ... An odd power will give you only the positive solution here. )
Once you have the value of y, you get a unique value for x and hence you get a value for xy.

I got this problem wrong. Bunuel, can you explain if I approached it the wrong way.

I factored out 200, which = \(2^3 * 5^2\).

After this its obvious \(xy = 10\)

What am i missing here?

Check my post above: if-x-2y-3-200-what-is-xy-1-y-is-an-integer-2-x-y-2-5-b-92486.html#p713396

First of all \(x^2y^3=2^3*5^2\) has infinitely many solutions for \(x\) and \(y\). For ANY (nonzero) value of \(x\) there exist some \(y\) to satisfy \(x^2y^3=200\) and vise-versa. For example \(x=1\) and \(y=\sqrt[3]{200}\), or \(x=10\) and \(y=\sqrt[3]{2}\), ... As you can see it's not necessary for \(x\) and \(y\) to be integers (3 and 2) to satisfy the given equation.

As for (1): \(y\) is an integer. \(y\) can be any positive integer and there will be two values of \(x\) for each \(y\) to satisfy \(x^2y^3=200\). For example \(y=10\) --> \(x^2y^3=x^2*1000=200\) --> \(x=\frac{1}{\sqrt{5}}\) or \(x=-\frac{1}{\sqrt{5}}\). So there are infinite values of \(xy\). Not sufficient.

Hope it's clear.

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